Chapter 9: Sequences and Series Exercise 9.3

Given G.P. is 5/2, 5/4, 5/8, ………

Here, a = First term = 5/2

r = Common ratio = (5/4)/(5/2) = ½

Thus, the 20^th term and n^th term

Given,

The common ratio of the G.P., r = 2

And, let a be the first term of the G.P.

Now,

a₈ = ar ^8–1 = ar⁷

ar⁷ = 192

a(2)⁷ = 192

a(2)⁷ = (2)⁶ (3)

Let’s take a to be the first term and r to be the common ratio of the G.P.

Then according to the question, we have

a₅ = a r^5–1 = a r⁴ = p … (i)

a₈ = a r^8–1 = a r⁷ = q … (ii)

a₁₁ = a r^11–1 = a r¹⁰ = s … (iii)

Dividing equation (ii) by (i), we get

Let’s consider a to be the first term and r to be the common ratio of the G.P.

Given, a = –3

And we know that,

a_n = ar^n–1

So, a₄ = ar³ = (–3) r³

a₂ = a r¹ = (–3) r

Then from the question, we have

(–3) r³ = [(–3) r]²

⇒ –3r³ = 9 r²

⇒ r = –3

a₇ = a r ^7–1 = a r⁶ = (–3) (–3)⁶ = – (3)⁷ = –2187

Therefore, the seventh term of the G.P. is –2187.

(a) The given sequence, 2, 2√2, 4,…

We have,

a = 2 and r = 2√2/2 = √2

Taking the n^th term of this sequence as 128, we have

Therefore, the 13^th term of the given sequence is 128.

(ii) Given the sequence, √3, 3, 3√3,…

We have,

a = √3 and r = 3/√3 = √3

Taking the n^th term of this sequence to be 729, we have

Therefore, the 12^th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, …

a = 1/3 and r = (1/9)/(1/3) = 1/3

Taking the n^th term of this sequence to be 1/19683, we have

Therefore, the 9^th term of the given sequence is 1/19683.

The given numbers are -2/7, x, -7/2.

Common ratio = x/(-2/7) = -7x/2

Also, common ratio = (-7/2)/x = -7/2x

Therefore, for x = ± 1, the given numbers will be in G.P.

Given G.P., 0.15, 0.015, 0.00015, …

Here, a = 0.15 and r = 0.015/0.15 = 0.1 

The given G.P is √7, √21, 3√7, ….

Here,

a = √7 and

The given G.P. is 1, -a, a², -a³ ….

Here, the first term = a₁ = 1

And the common ratio = r = – a

We know that,

Given G.P. is x³, x⁵, x⁷, …

Here, we have a = x³ and r = x⁵/x³ = x²